2. Your field 1 means and variances: space, time, spacetime
2. Field1 means and variances: space, time, spacetime
The GRAND (space+time) mean of my x and y: 1.0110e+003 hPa and27.5744 degC
The GRAND (space+time) varianceof my x and y: 2.3720 (hPa)^2 and 2.9326 (degC)^2
The GRAND standard deviations are: 1.5402 hPa and1.7125 degC
The SPATIAL variance of my TIME MEAN longitude section is: 1.1617 (hPa)^2 and 1.8427 (degC)^2
The TEMPORAL variance of my LONGITUDE MEAN time series is: 0.5488 (hPa)^2 and 0.3616 (degC)^2
3. Further decomposition of spacetime variations: Mean seasonal cycle and 'climate anomaly'
1) Confirm that the 20-year mean of the anomalies as defined above is 0. Write math (on paper, for yourself) that proves it/ shows why. Yes. Mean(anomx)=0,
2) Is the spatial mean of the climate anomalies (as defined above) 0? Is it the same as the time series of the spatial mean of the raw data? Or is it a new object? Yes, it is zero. It is definitely not the same as the time series of the spatial mean of the raw data. For one thing, the mean of the anomalies is zero whereas the mean of the raw data is not.
3) My CLIMATOLOGICAL ANNUAL CYCLE has variance: _(units). var( climx12(:),1)print, var_n(climx12) SLP: 1.8683 SST: 2.4852
4) My INTERANNUAL ANOMALY ARRAYS has variance: _(units). var( anomx(:),1)print, var_n(anomx) SLP: 0.5037 SST: 0.4473
5) Fill out a variance decomposition table for field 1: feel free to add columns if you can define other parts. It may help you to look at an example:
#
SLP
Uwnd (m/s)
a) total variance of x
2.3720
2.9326
b) purely spatial (variance of TIME mean at each lon)
1.1617
1.8427
c) temporal anomalies (x minus its TIME mean at each lon)
1.2103
1.0899
d) purely temporal (variance of LON mean at each time)
0.5488
0.3616
e) spatial anomalies (x minus its LON mean at each time)
1.8232
2.571
f) remove both means (space-time variability)
2.3720
2.9326
g) mean seasonal cycle
1.8683
2.4852
h) deseasonalized anomalies
0.5037
0.4473
i) variance of longitudinal mean of part h)
0.16
1.025
j) h minus i (anomalies from both space and monthly-climatological means)
0.34
.3448
A=G+H ie it checks out.
4. Further decomposition of anomx by scale (using rebinning).
There is a small gradient in the variance since there are small scale factors in space and time.
For such a small scale factor fixed in time, increasing the scale of the longitude does not change significantly change the gradient.
5. Scatter plot, correlation and covariance, regression-explained varianceBased on your data fields (which you've seen pictures of), make subsets of your 2 variables x and y and make a scatter plot of these showing the strongest (positive or negative) correlation of one field with the other you can find. The subset might simply be all (x,t) values if your fields are very similar (olr, precip), or maybe the 240 time values at one longitude, or 144 longitudinal values in the time mean, or time series at different longitudes if some variability is offset in your two fields (like pressure and wind).
Now consider the covariance and correlation of the two subset arrays entering your scatterplot.
rho= 1.0000 -0.7840
-0.7840 1.0000
standard deviation of slp at 190 longitude: 2.0855
standard deviation of sst at 190 longitude: 2.0085
2) What fraction of the variance of y can be 'explained' by linear regression on x (y = mx + b)? How does this relate to rho? How much y variance is explained? (variance: with units of y squared) What is m? Use the value from rho squared*total variance to get explained variance:
The total variance of SST at 190 longitude is 4.0342. The explained variance is 2.4533, which is 61% of the total variance.
The total variance of SLP at 190 longitude is 4.3292. The explained variance is 2.6460, which is 60% of the total variance.
This exercise is well explained on this page: http://www.mathworks.com/help/techdoc/data_analysis/f1-5937.html
Since I use matlab, polyfit(x,y,1) gives the y=mx+b formula. I got m=-0.7551
y=-0.7551*x+792
3) Now add uncorrelated (random) noise with variance 1 to one of your variables. This might be like observation error.
The variance of SST at 190 longitude with the noise is 5.3544, larger than the variance without noise (4.0342).
The value of rho is smaller, with a value of -0.6637.
The linear fit becomes y=-0.7364*x + 773
The total variance of SLP at 190 longitude with noise is 5.3544. The explained variance is 2.3586, which is 44% of the total variance. Predictably, adding
noise results in less explained variance. 6. Lagged correlation, covariance, and cross-covariance: hey let's compute all vs. all
2. Your field 1 means and variances: space, time, spacetime
2. Field1 means and variances: space, time, spacetime
3. Further decomposition of spacetime variations: Mean seasonal cycle and 'climate anomaly'
1) Confirm that the 20-year mean of the anomalies as defined above is 0. Write math (on paper, for yourself) that proves it/ shows why.
Yes. Mean(anomx)=0,
2) Is the spatial mean of the climate anomalies (as defined above) 0? Is it the same as the time series of the spatial mean of the raw data? Or is it a new object?
Yes, it is zero. It is definitely not the same as the time series of the spatial mean of the raw data. For one thing, the mean of the anomalies is zero whereas the mean of the raw data is not.
3) My CLIMATOLOGICAL ANNUAL CYCLE has variance: _(units). var( climx12(:),1) print, var_n(climx12)
SLP: 1.8683
SST: 2.4852
4) My INTERANNUAL ANOMALY ARRAYS has variance: _(units). var( anomx(:),1) print, var_n(anomx)
SLP: 0.5037
SST: 0.4473
5) Fill out a variance decomposition table for field 1: feel free to add columns if you can define other parts. It may help you to look at an example:
4. Further decomposition of anomx by scale (using rebinning).
variance_by_scalefactor =
0.5037 0.5022 0.4980 0.4884 0.4613 0.2811
0.3709 0.3698 0.3666 0.3592 0.3375 0.1861
0.2997 0.2988 0.2964 0.2903 0.2721 0.1388
0.2295 0.2287 0.2268 0.2221 0.2075 0.0951
0.1421 0.1416 0.1403 0.1371 0.1271 0.0536
0.0486 0.0484 0.0479 0.0464 0.0415 0.0142
There is a small gradient in the variance since there are small scale factors in space and time.
For such a small scale factor fixed in time, increasing the scale of the longitude does not change significantly change the gradient.
5. Scatter plot, correlation and covariance, regression-explained varianceBased on your data fields (which you've seen pictures of), make subsets of your 2 variables x and y and make a scatter plot of these showing the strongest (positive or negative) correlation of one field with the other you can find. The subset might simply be all (x,t) values if your fields are very similar (olr, precip), or maybe the 240 time values at one longitude, or 144 longitudinal values in the time mean, or time series at different longitudes if some variability is offset in your two fields (like pressure and wind).
Now consider the covariance and correlation of the two subset arrays entering your scatterplot.
rho= 1.0000 -0.7840
-0.7840 1.0000
standard deviation of slp at 190 longitude: 2.0855
standard deviation of sst at 190 longitude: 2.0085
2) What fraction of the variance of y can be 'explained' by linear regression on x (y = mx + b)? How does this relate to rho? How much y variance is explained? (variance: with units of y squared) What is m?
Use the value from rho squared*total variance to get explained variance:
The total variance of SST at 190 longitude is 4.0342. The explained variance is 2.4533, which is 61% of the total variance.
The total variance of SLP at 190 longitude is 4.3292. The explained variance is 2.6460, which is 60% of the total variance.
This exercise is well explained on this page: http://www.mathworks.com/help/techdoc/data_analysis/f1-5937.html
Since I use matlab, polyfit(x,y,1) gives the y=mx+b formula. I got m=-0.7551
y=-0.7551*x+792
3) Now add uncorrelated (random) noise with variance 1 to one of your variables. This might be like observation error.
The variance of SST at 190 longitude with the noise is 5.3544, larger than the variance without noise (4.0342).
The value of rho is smaller, with a value of -0.6637.
The linear fit becomes y=-0.7364*x + 773
The total variance of SLP at 190 longitude with noise is 5.3544. The explained variance is 2.3586, which is 44% of the total variance. Predictably, adding
noise results in less explained variance.
6. Lagged correlation, covariance, and cross-covariance: hey let's compute all vs. all